CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids NEET JEE Questions Set 03

Single Correct Answer

Question 1. The IUPAC name of the following compound is
H₃C–CH(CH₃)–CH(OH)–CH₂–COOH (JEE Main 2019)
(a) 2-methyl-3-hydroxypentan-5-oic acid
(b) 4, 4-dimethyl-3-hydroxybutanoic acid
(c) 4-methyl-3-hydroxypentanoic acid
(d) 3-hydroxy-4-methylpentanoic acid
Answer:
Ans. (d)
The COOH carbon is C-1 of the longest chain (5 C, pentanoic acid); OH is at C-3 and CH₃ at C-4.
Prefixes are written in alphabetical order (hydroxy before methyl): 3-hydroxy-4-methylpentanoic acid.

Teacher's Note:
1. Option (c) has the right locants but the wrong order; substituent prefixes must be alphabetical.

Question 2. RMgX + CO₂ dry
⟶
ether Y H₃O⁺
⟶ RCOOH
What is Y in the above reaction ? (NEET 2022)
(a) RCOMgX
(b) RCOOX
(c) (RCOO)₂Mg
(d) RCOOMgX
Answer:
Ans. (d)
R–MgX + O=C=O → R–COOMgX; the R group adds to the carbon of CO₂.
R–COOMgX H₃O⁺
⟶ R–COOH + Mg(OH)X

Teacher's Note:
1. Carbonation of a Grignard reagent adds one carbon: the acid has one C more than the alkyl halide.

Question 3. R–CH₂–CH₂OH can be converted into RCH₂CH₂COOH. The correct sequence of reagents is
(a) PBr₃, KCN, H⁺
(b) PBr₃, KCN, H₂
(c) KCN, H⁺
(d) HCN, PBr₃, H⁺
Answer:
Ans. (a)
RCH₂CH₂OH PBr₃
⟶ RCH₂CH₂Br KCN
⟶ RCH₂CH₂CN H⁺
⟶ RCH₂CH₂COOH

Teacher's Note:
1. OH is a poor leaving group, so it must be changed into Br before CN⁻ can displace it; acid hydrolysis then turns CN into COOH.

Question 4. The acid D obtained through the following sequence of reactions is :
C₂H₅Br Alc. KOH
⟶ A Br₂
⟶
CCl₄ B KCN
⟶
(excess) C H₃O⁺
⟶ D
(a) succinic acid
(b) malonic acid
(c) maleic acid
(d) oxalic acid.
Answer:
Ans. (a)
C₂H₅Br Alc. KOH
⟶ CH₂=CH₂ Br₂/CCl₄
⟶ CH₂Br–CH₂Br KCN
⟶ NCCH₂–CH₂CN H₃O⁺
⟶ HOOCCH₂–CH₂COOH (Succinic acid)

Teacher's Note:
1. Each CN group adds one carbon, so a two-carbon dibromide gives a four-carbon diacid.

Question 5. Grignard reagents and organolithium compounds on addition to dry ice separately, followed by hydrolysis gives (AIIMS 2013)
(a) ketones and carboxylic acids respectively
(b) carboxylic acids and ketones respectively
(c) only carboxylic acids
(d) only ketones
Answer:
Ans. (b)
Grignard reagents (RMgX) react with dry ice followed by hydrolysis gives carboxylic acids.
R–MgX + O=C=O → [R–CO–OMgX] H⁺/H₂O
⟶
−Mg(OH)X R–CO–OH (Carboxylic acids)
Organolithium compounds being more nucleophilic add to C=O of addition product (I) to give addition product (II) which upon subsequent hydrolysis gives a ketone.
R–Li + O=C=O → [R–CO–OLi] (I) R–Li
⟶ [R₂C(OLi)₂] (II) H⁺/H₂O
⟶ [R₂C(OH)₂] −H₂O
⟶ R–CO–R (Ketone)

Teacher's Note:
1. The carboxylate salt RCOOMgX is not attacked further by RMgX, but the more reactive RLi adds again to give a gem-diolate and finally a ketone.

Question 6. Which of the following reactions will not yield propionic acid ? (JEE Main 2021)
(a) CH₃CH₂CH₃ + KMnO₄ (Heat), OH⁻/H₃O⁺
(b) CH₃CH₂CCl₃ + OH⁻/H₃O⁺
(c) CH₃CH₂COCH₃ + OI⁻/H₃O⁺
(d) CH₃CH₂Br + Mg, CO₂, dry ether/H₃O⁺
Answer:
Ans. (a)
Alkanes are not oxidised by hot alkaline KMnO₄, therefore, propane is not oxidised to form propionic acid. However, all other reactions give propionic acid.
Hence, option (a) is correct.

Teacher's Note:
1. A CCl₃ group hydrolyses to COOH, a methyl ketone gives the acid by the haloform reaction, and RMgX + CO₂ adds one carbon.

Question 7. In the following conversion, the intermediate (A) is
(CH₃)₃CLi
⟶ A (i) CO₂
⟶
(ii) H₃O⁺ (JEE Main 2022)
(a)
(b)
(c)
(d)
Answer:
Ans. (b)
tert-Butyllithium first removes the acidic phenolic H (OH → OLi) and then exchanges Br for Li, giving the dilithio compound (A), o-LiC₆H₄OLi.
A (i) CO₂
⟶
(ii) H₃O⁺ o-HOC₆H₄COOH (salicylic acid). Thus, option (b) is correct.

Teacher's Note:
1. An organolithium reagent is a very strong base, so any O–H present is deprotonated before the C–Li bond is formed.

Question 8. Find out the major product for the following reaction :
(i) I₂/NaHCO₃
⟶
(ii) Pyridine, Δ (JEE Main 2022)
(a)
(b)
(c)
(d)
Answer:
Ans. (c)
I₂ adds to the C=C double bond to form a cyclic iodonium ion; with NaHCO₃ the carboxylate oxygen attacks the ring carbon of the iodonium ion from the back side, forming a five-membered iodolactone (iodolactonisation).
On heating with pyridine, HI is eliminated (the H from the CH₂ next to C=O and the I from the ring-fusion carbon), giving the α,β-unsaturated lactone.
Thus, option (c) is correct.

Teacher's Note:
1. Iodolactonisation closes the ring that has five atoms (O, C=O, CH₂ and two ring carbons); elimination of HI then puts the C=C in conjugation with the lactone C=O.

Question 9. What is the product of following reaction ?
Hex-3-ynal (i) NaBH₄
⟶
(ii) PBr₃, (iii) Mg/ether, (iv) CO₂/H₃O⁺ ? (JEE Main 2020)
(a)
(b)
(c)
(d)
Answer:
Ans. (b)
CH₃CH₂C≡CCH₂CHO (Hex-3-ynal) (i) NaBH₄
⟶
CHO → CH₂OH CH₃CH₂C≡CCH₂CH₂OH
(ii) PBr₃
⟶
CH₂OH → CH₂Br CH₃CH₂C≡CCH₂CH₂Br (iii) Mg/ether
⟶ CH₃CH₂C≡CCH₂CH₂MgBr (Grignard reagent)
(iv) CO₂
⟶ CH₃CH₂C≡CCH₂CH₂COO⁻MgBr⁺ H₃O⁺
⟶ CH₃CH₂C≡CCH₂CH₂COOH
Thus, option (b) is correct.

Teacher's Note:
1. NaBH₄ reduces only the CHO group; the C≡C bond survives every step, and CO₂ adds one carbon at the end.

Question 10. Decarboxylation of all six possible forms of diaminobenzoic acids, C₆H₃(NH₂)₂COOH yields three products A, B and C. Three acids give a product ‘A’, two acids give a product ‘B’ and one acid gives a product ‘C’. The melting point of product ‘C’ is (JEE Main 2022)
(a) 65°C
(b) 90°C
(c) 104°C
(d) 142°C
Answer:
Ans. (d)
Six diaminobenzoic acids are the 2,3-, 2,4-, 2,5-, 2,6-, 3,4- and 3,5-isomers.
On heating (−CO₂), the 2,4-, 2,6- and 3,5-isomers (I, II, III) give m-diaminobenzene (A); the 2,3- and 3,4-isomers (IV, V) give o-diaminobenzene (B); and only the 2,5-isomer (VI) gives p-diaminobenzene (C).
Out of three diaminobenzenes (A, B, C), only p-diaminobenzene (C) being symmetrical fits closely in the crystal lattice and hence it has the highest m.p. 142°C.
Thus, option (d) is correct.

Teacher's Note:
1. Among o-, m- and p-isomers the para isomer is the most symmetrical and usually has the highest melting point.

Question 11. Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their (NEET 2018)
(a) formation of intramolecular H-bonding
(b) formation of carboxylate ion
(c) more extensive association of carboxylic acid via van der Waals’ forces of attraction
(d) formation of intermolecular H-bonding
Answer:
Ans. (d)
Formation of intermolecular H-bonding.
Carboxylic acid molecules are held together as dimers (and larger aggregates) by strong intermolecular hydrogen bonds, which persist even in the vapour phase.

Teacher's Note:
1. The O–H of COOH is more polar than that of an alcohol, so its H-bonds are stronger; that is why acids boil higher than alcohols.

Question 12. What will be the correct decreasing order of acid strength of the hydroxybenzoic acids ? (J & K CET 2018)
(a) p-Hydroxybenzoic acid > benzoic acid > m-hydroxybenzoic acid > o-hydroxybenzoic acid
(b) o-Hydroxybenzoic acid > m-hydroxybenzoic acid > benzoic acid > p-hydroxybenzoic acid
(c) o-Hydroxybenzoic acid > benzoic acid > m-hydroxybenzoic acid > p-hydroxybenzoic acid
(d) m-Hydroxybenzoic acid > benzoic acid > o-hydroxybenzoic acid > p-hydroxybenzoic acid
Answer:
Ans. (b)
o-Hydroxybenzoic acid is the strongest because its carboxylate ion is stabilised by intramolecular H-bonding with the ortho OH (ortho effect).
At the m-position OH exerts only its −I effect and increases the acid strength slightly above that of benzoic acid, while at the p-position its +R effect decreases the acid strength. Thus, option (b) is correct.

Teacher's Note:
1. Almost any ortho substituent makes benzoic acid stronger (ortho effect); for m/p compare −I against +R.

Question 13. The correct order of acid strength of the following carboxylic acids is
I. HC≡C–COOH   II. CH₂=CH–COOH   III. p-MeO–C₆H₄–COOH   IV. CH₃CH₂COOH (JEE Advanced 2019)
(a) I > II > III > IV
(b) II > I > IV > III
(c) I > III > II > IV
(d) III > II > I > IV
Answer:
Ans. (a)
As the electronegativity of the α-carbon decreases with the change of hybridization of carbon from sp–sp²–sp³, the acid strength decreases.
Although α-carbon of both acid (II) and acid (III) are sp²-hybridized, but the electron-donating CH₃O group decreases the acidity of acid (III) relative to acid (II). Thus, the overall acidity decreases in the order : I > II > III > IV.

Teacher's Note:
1. More s-character on the carbon next to COOH means a more electronegative carbon and a stronger acid: sp > sp² > sp³.

Question 14. The correct order of strengths of the carboxylic acids
is (NEET Phase-II 2016)
(a) I > II > III
(b) II > III > I
(c) III > II > I
(d) II > I > III
Answer:
Ans. (b)
Presence of electron-withdrawing group (i.e., O in the present case) increases the acidic strength of carboxylic acids. Therefore, carboxylic acids (II and III) are more acidic than carboxylic acid (I).
Further since −I-effect decreases with distance, therefore, carboxylic (II) (in which O is nearer the COOH group) is more acidic than carboxylic acid (III) (in which O is farther from the COOH group).
Combining both the trends, the acidic strength of the three acids decreases in the order : II > III > I, i.e., option (b) is correct.

Teacher's Note:
1. The −I effect falls off quickly with the number of bonds, so the ring O helps most when it is closest to the COOH group.

Question 15. The increasing order of the reactivity of the following with LiAlH₄ is
(A) C₂H₅CONH₂   (B) C₂H₅COOCH₃   (C) C₂H₅COCl   (D) C₂H₅CO–O–COC₂H₅ (JEE Main 2019)
(a) (A) < (B) < (D) < (C)
(b) (B) < (A) < (D) < (C)
(c) (B) < (A) < (C) < (D)
(d) (A) < (B) < (C) < (D)
Answer:
Ans. (a)
Reactivity of the acyl derivative increases as the basicity of the leaving group (i.e., NH₂⁻ > ⁻OCH₃ > ⁻O–CO–C₂H₅ > Cl⁻) decreases. In other words, reactivity increases in the order : A < B < D < C.

Teacher's Note:
1. The weaker the base that leaves, the more reactive the acyl derivative: acid chloride > anhydride > ester > amide.

Question 16. Which of the following acid will form an (a) anhydride on heating and (b) acid imide on heating with ammonia ? (NEET 2020)
(a)
(b)
(c)
(d)
Answer:
Ans. (d)
o-Substituted dicarboxylic acids on heating undergo dehydration to form anhydrides and imides on strong heating with NH₃.
Phthalic acid Δ
⟶
−H₂O Phthalic anhydride; Phthalic acid NH₃/Δ
⟶
−2H₂O Phthalimide. Thus, option (d) is correct.

Teacher's Note:
1. A cyclic anhydride or imide needs the two COOH groups close enough to form a five- or six-membered ring, which in benzene means the ortho positions.

Question 17. An organic compound (A) (molecular formula, C₆H₁₂O₂) was hydrolysed with dilute H₂SO₄ to give a carboxyl acid (B) and alcohol (C). ‘C’ gives white turbidity immediately when treated with anhydrous ZnCl₂ and conc. HCl. The organic compound (A) is (JEE Main 2020)
(a) CH₃COOCH₂CH₂CH₂CH₃
(b) CH₃COOCH₂CH(CH₃)₂
(c) CH₃COOCH(CH₃)CH₂CH₃
(d) CH₃COOC(CH₃)₃
Answer:
Ans. (d)
Since alcohol ‘C’ immediately gives white turbidity with Lucas reagent (anhyd. ZnCl₂ + conc. HCl), therefore, alcohol (C) must be tertiary butyl alcohol.
Therefore, ester (A) must contain a tertiary butyl group and hence option (d) is correct.

Teacher's Note:
1. Lucas test: immediate turbidity for 3° alcohols, about 5 minutes for 2°, and none at room temperature for 1°.

Question 18. Which of the following esters gets hydrolysed most easily under alkaline conditions ? (RE-AIPMT 2015)
(a) p-CH₃O–C₆H₄–OCOCH₃
(b) C₆H₅–OCOCH₃
(c) p-Cl–C₆H₄–OCOCH₃
(d) p-O₂N–C₆H₄–OCOCH₃
Answer:
Ans. (d)
Due to powerful −I and −R-effect of NO₂, the +ve charge on the C=O group of the ester increases.
As a result, nucleophilic attack by the OH⁻ occurs readily and hence the ester gets hydrolysed easily.

Teacher's Note:
1. An electron-withdrawing group on the phenol part also makes the phenoxide a better (weaker-base) leaving group.

Question 19. Major product of the following reaction is :
CH₂=CH–CO–O–CH₂–CH=CH₂ 2 HBr
⟶ (JEE Main 2022)
(a) CH₃–CHBr–CO–O–CH₂CH₂CH₂Br
(b) CH₃–CHBr–CO–O–CH₂–CHBr–CH₃
(c) BrCH₂CH₂–CO–O–CH₂CH₂CH₂Br
(d) BrCH₂CH₂–CO–O–CH₂–CHBr–CH₃
Answer:
Ans. (d)
Due to electron-withdrawing resonance effect of C=O, Br⁻ attacks at the β-carbon of the α,β-unsaturated ester and due to −I-effect of oxygen atom Br⁻ attacks at the β′ carbon giving (d) as the final product.

Teacher's Note:
1. HBr adds to an α,β-unsaturated carbonyl compound in a conjugate (anti-Markovnikov looking) way, but to an isolated C=C by Markovnikov’s rule.

Question 20. The reagent(s) required for the following conversion are
→ (JEE Main (online) 2018)
(a) (i) LiAlH₄, (ii) H₃O⁺
(b) (i) B₂H₆, (ii) DIBAL-H, (iii) H₃O⁺
(c) (i) B₂H₆, (ii) SnCl₂/HCl, (iii) H₃O⁺
(d) (i) NaBH₄, (ii) Raney Ni/H₂, (iii) H₃O⁺
Answer:
Ans. (c)
Since DIBAL-H also reduces ester, acids and cyanides to –CHO group, therefore, B₂H₆ is used to reduce CO₂H to CH₂OH.
(i) B₂H₆ reduces –COOH → CH₂OH; (ii) SnCl₂/HCl reduces CN to CHO; (iii) H₃O⁺ hydrolyses CO₂Et to CO₂H.
Thus, option (c) is correct.

Teacher's Note:
1. B₂H₆ reduces COOH faster than esters or nitriles; Stephen reduction (SnCl₂/HCl, then H₃O⁺) turns CN into CHO.

Question 21. When CH₂=CH–COOH is reduced with LiAlH₄, the compound obtained will be
(a) CH₃–CH₂–CHO
(b) CH₃–CH₂–COOH
(c) CH₂=CH–CH₂OH
(d) CH₃–CH₂–CH₂OH
Answer:
Ans. (c)
LiAlH₄ reduces only the –COOH without reducing the double bond.
CH₂=CH–COOH LiAlH₄
⟶ CH₂=CH–CH₂OH

Teacher's Note:
1. LiAlH₄ delivers H⁻ to polar C=O groups; isolated C=C bonds are not reduced by it.

Question 22. A compound undergoes the following sequence of reactions :
C₃H₅N Hydrolysis
⟶ C₃H₆O₂ (A) Cl₂/P
⟶ C₃H₅O₂Cl (B) NH₃
⟶ C₃H₇NO₂ (C)
The compound C is
(a) 1-nitropropane
(b) 2-nitropropane
(c) 2-aminopropanoic acid
(d) 2-hydroxypropanamide.
Answer:
Ans. (c)
CH₃CH₂CN (Propanenitrile, C₃H₅N) Hydrolysis
⟶ CH₃CH₂COOH, C₃H₆O₂ (A) Cl₂/P
⟶ CH₃CHClCOOH, C₃H₅O₂Cl (B)
NH₃
⟶ CH₃–CHNH₂–COOH, 2-Aminopropanoic acid (C), C₃H₇NO₂

Teacher's Note:
1. HVZ puts Cl on the α-carbon, and NH₃ then replaces it by NH₂; this is a route to α-amino acids.

Question 23. Which of the following acids on heating loses a molecule of H₂O to form an α,β-unsaturated acid ?
(a) CH₃CHOHCOOH
(b) HOCH₂COOH
(c) CH₃CHOHCH₂COOH
(d) HOCH₂CH₂CH₂COOH.
Answer:
Ans. (c)
β-Hydroxy acids on heating lose a molecule of H₂O to form α,β-unsaturated acids, i.e., option (c) is correct.
CH₃CH(OH)CH₂COOH heat
⟶ CH₃CH=CHCOOH + H₂O

Teacher's Note:
1. On heating, α-hydroxy acids form lactides, β-hydroxy acids lose water to give α,β-unsaturated acids, and γ-hydroxy acids form lactones.

Question 24. Which of the following carboxylic acids undergoes decarboxylation easily ? (AIIMS 2014)
(a) C₆H₅–CO–CH₂–COOH
(b) C₆H₅–CO–COOH
(c) C₆H₅–CH(OH)–COOH
(d) C₆H₅–CH(NH₂)–COOH
Answer:
Ans. (a)
β-Keto acids on heating easily undergo decarboxylation via a six-membered cyclic transition state.
C₆H₅–CO–CH₂–COOH −CO₂
⟶ [C₆H₅–C(OH)=CH₂] Tautomerises
⟶ C₆H₅–CO–CH₃

Teacher's Note:
1. Only a β-keto acid can form the six-membered cyclic transition state; an α-keto acid such as C₆H₅COCOOH cannot.

Question 25. The compound that undergoes decarboxylation most readily under mild condition is (IIT 2012)
(a)
(b)
(c)
(d)
Answer:
Ans. (b)
β-Ketoacids on heating readily undergo decarboxylation.
2-Oxocyclohexanecarboxylic acid Δ
⟶
−CO₂ [cyclohex-1-en-1-ol] Tautomerises
⟶ Cyclohexanone

Teacher's Note:
1. Look for a C=O group on the carbon β to the COOH carbon; in (d) the ketone is γ to COOH and does not help.

Question 26. The major product in the reaction of N-phenylbenzamide with Br₂/Fe is (JEE Main 2021)
(a)
(b)
(c)
(d)
Answer:
Ans. (b)
Due to electron-donating effect of the NH group, bromination occurs at less hindered p-position in ring attached to it (i.e., NH group).
C₆H₅–CO–NH–C₆H₅ Br₂/Fe
⟶ C₆H₅–CO–NH–C₆H₄–Br (p)

Teacher's Note:
1. In an anilide the ring on N is activated (o/p-directing), while the ring on C=O is deactivated; the bulky amide group favours the para position.

Question 27. The structure of product C, formed by the following sequence of reactions is :
CH₃COOH + SOCl₂ → A Benzene
⟶
AlCl₃ B KCN
⟶
⁻OH C (JEE Main 2021)
(a) C₆H₅–CH(CH₃)–COOH
(b) C₆H₅–C(OH)(CN)–CH₃
(c) C₆H₅–CH₂CH₂CN
(d)
Answer:
Ans. (b)
CH₃COOH SOCl₂
⟶
−SO₂, −HCl CH₃COCl (A) Benzene, AlCl₃
⟶
(F.C. reaction) C₆H₅COCH₃ (Acetophenone, B) KCN
⟶
⁻OH C₆H₅C(OH)(CN)CH₃ (C)
Thus, option (b) is correct.

Teacher's Note:
1. KCN with a base supplies CN⁻ which adds to the ketone C=O to form a cyanohydrin; it does not substitute on the ring.

Question 28. The structures of A and B formed in the following reaction are (Ph = C₆H₅)
Benzene + succinic anhydride AlCl₃ (2 eq.)
⟶ A Zn/Hg
⟶
HCl B (JEE Main 2021)
(a) A = Ph–CO–CH₂CH₂–COOH, B = Ph–CH₂CH₂CH₂–COOH
(b) A = Ph–CO–CH₂CH₂–COOH, B = Ph–CH₂CH₂CH₂CH₂–OH
(c) A = Ph–CO–CH₂CH₃, B = Ph–CH(OH)–CH₂CH₃
(d) A = Ph–CO–CH₂CH₃, B = Ph–CH₂CH₂CH₃
Answer:
Ans. (a)
C₆H₆ + (CH₂CO)₂O AlCl₃ (2 eq.)
⟶
(F.C. reaction) Ph–CO–CH₂CH₂–COOH (A) Zn/Hg/HCl
⟶
(Clemmensen reduction) Ph–CH₂CH₂CH₂–COOH (B)
Thus, option (a) is correct.

Teacher's Note:
1. Clemmensen reduction converts the ketone C=O to CH₂ but leaves the COOH group unchanged.

One or More Correct Answers

Question 29. The reaction(s) leading to the formation of 1, 3, 5-trimethylbenzene is (are) (JEE Advanced 2018)
(a) CH₃COCH₃ conc. H₂SO₄
⟶
Δ
(b) Me–C≡C–H heated iron tube
⟶
873 K
(c) (i) Br₂/NaOH (ii) H₃O⁺
⟶
(iii) soda-lime, Δ
(d) Zn/Hg, HCl
⟶
Answer:
Ans. (a, b, d)
Only option (c) does not give 1, 3, 5-trimethylbenzene while all the remaining three give.
(a) 3CH₃COCH₃ conc. H₂SO₄
⟶
−3H₂O 1, 3, 5-Trimethylbenzene
(b) 3 Me–C≡C–H heated iron tube
⟶
873 K 1, 3, 5-trimethylbenzene (cyclic trimerisation)
(c) 1,3,5-Triacetylbenzene (i) Br₂/NaOH
⟶
(Iodoform reaction) (−3CHBr₃) C₆H₃(CO₂Na)₃ H₃O⁺
⟶ C₆H₃(CO₂H)₃ Soda-lime, Δ
⟶
Decarboxylation (−3CO₂) Benzene
(d) Benzene-1,3,5-tricarbaldehyde Zn/Hg, HCl
⟶
(Clemmensen reduction) 1,3,5-trimethylbenzene

Teacher's Note:
1. In (c) the haloform reaction removes the methyl carbons as CHBr₃ and soda-lime removes the COOH carbons, so only the benzene ring is left.

Question 30. Which of the following reactions produce(s) propane as a major product ? (JEE Advanced 2019)
(a) CH₃CH₂CH₂COONa NaOH, CaO, Δ
⟶
(b) CH₃CH₂CH₂COONa + H₂O Electrolysis
⟶
(c) CH₃CH₂CH₂Cl Zn dil. HCl
⟶
(d) CH₃CHBrCH₂Br Zn
⟶
Answer:
Ans. (a, c)
(a) Decarboxylation with soda-lime removes CO₂: CH₃CH₂CH₂COONa + NaOH CaO, Δ
⟶ CH₃CH₂CH₃ + Na₂CO₃
(b) Kolbe electrolysis joins two propyl radicals: 2CH₃CH₂CH₂COONa + 2H₂O → CH₃(CH₂)₄CH₃ (hexane) + 2CO₂ + H₂ + 2NaOH
(c) Zn/dil. HCl reduces the alkyl chloride: CH₃CH₂CH₂Cl + 2[H] → CH₃CH₂CH₃ + HCl
(d) Zn removes two vicinal Br atoms (dehalogenation): CH₃CHBrCH₂Br + Zn → CH₃CH=CH₂ (propene) + ZnBr₂

Teacher's Note:
1. Soda-lime decarboxylation gives an alkane with one carbon less, whereas Kolbe electrolysis doubles the alkyl group.

Passage-Based Questions

Question 31. Comprehension 2: The acidic strength of saturated aliphatic carboxylic acids depends mainly upon the inductive effect of the substituent and its position w.r.t. the –COOH group. Whereas electron-donating substituents tend to decrease, electron-withdrawing substituents tend to increase the acid strength. The acidic strength of aromatic carboxylic acids, on the other hand, depends upon both the inductive and the resonance effect of the substituents.
Among the following, the strongest acid is
(a) CH₃COOH
(b) C₆H₅COOH
(c) m-CH₃OC₆H₄COOH
(d) p-CH₃OC₆H₄COOH
Answer:
Ans. (c)
Since in C₆H₅COOH, –COOH is attached to more electronegative sp²-carbon of the C₆H₅ ring while in CH₃COOH, it is attached to less electronegative sp³-carbon of CH₃ group, therefore, benzoic acid is a stronger acid than acetic acid.
Since methoxy group acts as an electron-withdrawing group at m-position but electron-donating group at p-position, therefore, m-methoxybenzoic acid is even more acidic than benzoic acid. Thus, option (c) is correct.

Teacher's Note:
1. At the meta position a methoxy group can act only through its −I effect, because its +R effect reaches only o- and p-positions.

Question 32. Comprehension 2: The acidic strength of saturated aliphatic carboxylic acids depends mainly upon the inductive effect of the substituent and its position w.r.t. the –COOH group. Whereas electron-donating substituents tend to decrease, electron-withdrawing substituents tend to increase the acid strength. The acidic strength of aromatic carboxylic acids, on the other hand, depends upon both the inductive and the resonance effect of the substituents.
Which of the following is obtained when 4-methylbenzenesulphonic acid is hydrolysed with excess of sodium acetate ?
(a) p-CH₃–C₆H₄–COO⁻Na⁺
(b) CH₃–C₆H₅ + SO₃
(c) p-CH₃–C₆H₄–SO₃⁻Na⁺ + CH₃COOH
(d) p-CH₃–C₆H₄–SO₂O.COCH₃ + NaOH
Answer:
Ans. (c)
4-Methylbenzenesulphonic acid is a stronger acid than acetic acid, therefore, it displaces acetic acid from sodium acetate. Thus,
4-Me–C₆H₄–SO₃H + CH₃COONa → 4-Me–C₆H₄SO₃⁻Na⁺ + CH₃COOH

Teacher's Note:
1. A stronger acid always displaces a weaker acid from its salt; sulphonic acids are far stronger than carboxylic acids.

Matrix Match Questions

Teacher's Note:
1. In cyanohydrin formation the carbonyl compound is the electrophile and CN⁻ the nucleophile, so both are ‘involved’.

Teacher's Note:
1. An NO₂ group ortho or para to a halogen activates the ring for nucleophilic substitution.

Integer and Numerical Answer Questions

Question 35. Amongst the following, total number of compounds soluble in sodium bicarbonate are : 2, 4, 6-trinitrophenol, benzoic acid, salicylic acid, acetyl chloride, acetic anhydride, trifluoroethanol, acetamide, benzenesulphonic acid.
Answer:
Ans. 6
Six compounds, i.e., 2, 4, 6-trinitrophenol, benzoic acid, salicylic acid, acetyl chloride, acetic anhydride and benzenesulphonic acid all decompose NaHCO₃ and hence dissolve in it.

Teacher's Note:
1. A compound dissolves in NaHCO₃ (with effervescence of CO₂) if it is, or is quickly hydrolysed to, an acid stronger than H₂CO₃; picric acid qualifies but ordinary phenols do not.

Question 36. How many β-ketoacids on heating undergo decarboxylation to give 2-methylcyclohexanone ?
Answer:
Ans. 2
Two. β-Ketoacids, i.e., 3-methyl-2-oxocyclohexanecarboxylic acid and 1-methyl-2-oxocyclohexanecarboxylic acid

Teacher's Note:
1. The COOH must be on an α-carbon of the ketone; the two α-carbons of 2-methylcyclohexanone give two different acids.

Question 37. The total number of carboxylic acid groups in the product P is
1. H₃O⁺
⟶
2. O₃, 3. H₂O₂ P (JEE Advanced 2013)
Answer:
Ans. 2
Two as shown below :
Hydrolysis (H₃O⁺) of the anhydride gives a dicarboxylic acid in which each COOH is β to a ring C=O group (a β-keto acid); on heating both lose CO₂ (Δ, −2CO₂, decarboxylation).
Oxidative ozonolysis (O₃/H₂O₂) of the ring C=C then gives two COOH groups (product A).
No. of –COOH group in the final product (A) is two.

Teacher's Note:
1. β-Keto acids formed during a reaction usually lose CO₂ under the acidic, warm conditions, so count COOH groups only after decarboxylation.

Question 38. In the following reaction sequence, the amount of D (in g) formed from one mole of anisole is ...........
Anisole Succinic anhydride
⟶
Anhyd. AlCl₃ A (75%) Zn-Hg/conc. HCl
⟶ B (70%) H₃PO₄
⟶ C (80%) NH₂OH.HCl
⟶
CH₃COONa D (95%)
Answer:
Ans. 76.21
Anisole (1 mole) + succinic anhydride Anhyd. AlCl₃
⟶
(F.C. acylation) 4-(4-Methoxyphenyl)-4-oxobutanoic acid (A) (75%)
A Zn-Hg/conc. HCl
⟶
(Clemmensen reduction) 4-(4-Methoxyphenyl)butanoic acid (B) (70%) H₃PO₄
⟶
(Cyclization) 7-Methoxy-1-tetralone (C) (80%)
C NH₂OH.HCl
⟶
CH₃COONa 7-Methoxy-1-tetralone oxime (D) (95%)
Amount of D (in moles) \( = 1 \times \frac{75}{100} \times \frac{70}{100} \times \frac{80}{100} \times \frac{95}{100} \) = 0.399
M.F. of D = C₁₁H₁₃NO₂; Mol. mass of D = 11 × 12 + 13 × 1 + 14 × 1 + 2 × 16 = 191 g mol⁻¹
∴ Amount of D (in g) = 0.399 × 191 = 76.209 = 76.21

Teacher's Note:
1. Multiply the fractional yields of all the steps to get the overall yield, then convert moles of the final product into grams.

Assertion–Reason Questions

Question 39. Assertion: Carboxylic acids contain a carbonyl group but do not give characteristic reactions of the carbonyl group.
Reason: Due to resonance, the electrophilic nature of the carboxyl carbon is greatly reduced as compared to the carbonyl carbon in aldehydes and ketones.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
In R–COOH the lone pair of the OH oxygen is delocalised onto the C=O group, so the carboxyl carbon is much less electrophilic and does not add reagents such as HCN, NaHSO₃, NH₂OH, etc. Reason explains the assertion.

Teacher's Note:
1. Acids, esters and amides all have resonance-stabilised C=O groups; they react by substitution rather than addition.

Question 40. Assertion: The pK_a of acetic acid is lower than that of phenol.
Reason: Phenoxide ion is more resonance stabilized than acetate ion.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct reason. Acetate ion is more resonance stabilized than phenoxide ion.
In acetate the negative charge is shared equally by two electronegative O atoms, whereas in phenoxide it is delocalised onto less electronegative ring carbons.

Teacher's Note:
1. Lower pK_a means stronger acid: acetic acid (4.76) is far stronger than phenol (10.0).

Question 41. Assertion: Benzoic acid is a weaker acid than formic acid.
Reason: Phenyl group when attached to carbonyl group becomes electron-donating.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
The phenyl group attached to the carboxyl group releases electrons by resonance (+R effect), which destabilises the benzoate ion relative to the formate ion; hence benzoic acid (pK_a 4.20) is weaker than formic acid (pK_a 3.75). Reason explains the assertion.

Teacher's Note:
1. Formic acid is the strongest of the simple monocarboxylic acids because H has no electron-releasing effect.

Question 42. Assertion: Although fluorine is more electronegative than chlorine, yet p-fluorobenzoic acid is a weaker acid than p-chlorobenzoic acid.
Reason: Due to matching sizes of 2p-orbitals of F and carbon, F has a stronger +R-effect than Cl.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
At the para position, the +R effect of the halogen opposes its −I effect. F (2p–2p overlap with carbon) has a much stronger +R effect than Cl (3p–2p overlap), which outweighs its stronger −I effect. Hence p-fluorobenzoic acid is the weaker acid. Reason explains the assertion.

Teacher's Note:
1. Orbital size match matters for resonance: 2p–2p overlap is effective, 3p–2p overlap is poor.

Question 43. Assertion: Malonic acid (HOOC–CH₂–COOH) does not form cyclic anhydride on heating.
Reason: It is like β-keto acid, on heating, it prefers to decarboxylate. (AIIMS 2014)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
A cyclic anhydride of malonic acid would be a strained four-membered ring. Like a β-keto acid, malonic acid on heating loses CO₂ through a six-membered cyclic transition state: HOOC–CH₂–COOH Δ
⟶ CH₃COOH + CO₂. Reason explains the assertion.

Teacher's Note:
1. Succinic and glutaric acids (five- and six-membered rings) form cyclic anhydrides; malonic acid decarboxylates instead.

Question 44. Assertion: Formic acid reduces Tollens’ reagent.
Reason: Compounds containing CHO group reduce Tollens’ reagent.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
Correct explanation. Formic acid (H–CO–OH) contains an aldehydic (CHO) group; it can be easily oxidised to carbonic acid which decomposes to form CO₂ and H₂O.

Teacher's Note:
1. Formic acid is the only common carboxylic acid that gives the Tollens’ and Fehling’s tests, because of its H–C=O part.

Question 45. Assertion: Acetamide has more polar >C=O group than in ethyl acetoacetate.
Reason: NH₂ is more electron-donating than OC₂H₅. (AIIMS 2007)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
Because NH₂ donates its lone pair to the C=O group more strongly (+R) than OC₂H₅, the contribution of the dipolar structure >C⁺–O⁻ is larger in acetamide; hence its C=O group is more polar. Reason explains the assertion.

Teacher's Note:
1. N is less electronegative than O, so it donates its lone pair more readily; amides have the most polar (and least reactive) C=O among acid derivatives.